Maple 2019 Questions and Posts

These are Posts and Questions associated with the product, Maple 2019

I have the following expression

I would like to know if is it possible avoid negative exponent goes to denominator when I use expand().

Thank you!

 

 

 

 

 

 

Maple always starts a new document with a 2D input style and Times New Roman font. How to change the boot style? (default).
Oliveira

I have to solve the equation rH''(r)+H'(r)+(rk^2-r^2*b^2/R^2)=0 where k, b, and R are real constant positive number, with condition H(R)=0 and H(1/R)=R to be solved into series of power. I know from the literature that xy''+y'+xy=0, can't be solved in terms of elementary function(see G.Nagy-ODE-November 29, 2017) that's why I'm interested in an approximate solution based on series, or any results as long as it satisfied the too condition H(R)=0 and H(1/R)=R of the real function H(r). 

Please advice!.

    I have encountered a peculiar behavior in Maple 2019 worksheets. I have attached a worksheet which illustrates a reversal of the coefficient and the blade in the following types of expression.  My Maple installation is set in options to use maple input and output (1D notation).  I am using build ID 1399874.

The procedure I was testing used the add function to expand a multivector over basis blades and coefficients represented by indexed names; for example
    add(a[indx[]]*e[indx[]], indx = indxes); where indxes:={[1],[2],[3],[1,2],[1,3],[1,4],[2,3],[2,4]};

The expected result
a[1]*e[1]+a[2]*e[2]+a[3]*e[3]+a[4]*e[1, 2]+a[5]*e[1, 3]+a[6]*e[1, 4]+a[7]*e[2, 3]+a[8]*e[2, 4] in 1D notation but instead I got

e[1]*xx[1]+e[2]*xx[2]+e[3]*xx[3]+e[1, 2]*xx[4]+e[1, 3]*xx[5]+e[1, 4]*xx[6]+ e[2, 3]*xx[7]+e[2, 4]*xx[8]

(Note and the different coefficient names illustrates the exchange is caused by sorting.)

Initially, I assumed that this resulted from the add command, but after I directly entered an indexed expression into the worksheet and obtained a similar result I realized it must be occurring during the output.  Apparently, before the expression is written, it is first sorted. I know this because, if the coefficient is named a,b,c,d ( less than e), the expression is not exchanged. In addition, if I convert the expression to 2D output, the expression is written in the normal order specified in the add command.

I also confirmed this behavior in Maple 2018.2 as well.  Hopefully, the example worksheet attached will illustrate this behavior.  Am I correct in assuming that when the code I am using in the worksheet with is incorporated into a module in an mpl file and loaded, this will not be an issue.

exchangeproblem.mw

Hello,

How I can extract coefficients from and by calculating determinant for Eigenvalue problem, the value of omega.

For more details please see attached PDF file.

Thanks so much.

eign.pdf

 

We solve Laplace's equation in the domain a < r and r < b, c < t and t < d
in polar coordinates subject to prescribed Dirichlet data.

Maple produces a solution in the form of an infinite sum,
but that solution fails to satisfy the boundary condition
on the domain's outer arc.  Is this a bug or am I missing
something?

restart;

kernelopts(version);

`Maple 2019.1, X86 64 LINUX, May 21 2019, Build ID 1399874`

with(plots):

pde := diff(u(r,t),r,r) + diff(u(r,t),r)/r + diff(u(r,t),t,t)/r^2 = 0;

diff(diff(u(r, t), r), r)+(diff(u(r, t), r))/r+(diff(diff(u(r, t), t), t))/r^2 = 0

a, b, c, d := 1, 2, Pi/6, Pi/2;

1, 2, (1/6)*Pi, (1/2)*Pi

bc := u(r,c)=c, u(r,d)=0, u(a,t)=0, u(b,t)=t;

u(r, (1/6)*Pi) = (1/6)*Pi, u(r, (1/2)*Pi) = 0, u(1, t) = 0, u(2, t) = t

We plot the boundary data on the domain's outer arc:

p1 := plots:-spacecurve([b*cos(t), b*sin(t), t], t=c..d, color=red, thickness=5);

Solve the PDE:

pdsol := pdsolve({pde, bc});

u(r, t) = Sum((1/6)*cos(3*signum(n1-1/4)*(-1+4*n1)*t)*(2*Pi*sin((1/2)*signum(n1-1/4)*Pi)*abs(n1-1/4)-6*Pi*sin((3/2)*signum(n1-1/4)*Pi)*abs(n1-1/4)+cos((3/2)*signum(n1-1/4)*Pi)-cos((1/2)*signum(n1-1/4)*Pi))*signum(n1-1/4)*8^(signum(n1-1/4)*(4*n1+1))*(r^((-3+12*n1)*signum(n1-1/4))-r^((3-12*n1)*signum(n1-1/4)))/(abs(n1-1/4)*Pi*(-1+4*n1)*(16777216^(signum(n1-1/4)*n1)-64^signum(n1-1/4))), n1 = 0 .. infinity)+Sum(-(1/3)*((-1)^n-1)*sin(n*Pi*ln(r)/ln(2))*(exp((1/6)*Pi*n*(Pi+6*t)/ln(2))-exp((1/6)*Pi*n*(7*Pi-6*t)/ln(2)))/(n*(exp((1/3)*n*Pi^2/ln(2))-exp(n*Pi^2/ln(2)))), n = 1 .. infinity)

Truncate the infinite sum at 20 terms, and plot the result:

eval(rhs(pdsol), infinity=20):
value(%):
p2 := plot3d([r*cos(t), r*sin(t), %], r=a..b, t=c..d);

Here is the combined plot of the solution and the boundary condition.
We see that the proposed solution completely misses the boundary condition.

plots:-display([p1,p2], orientation=[25,72,0]);


 

Download mw.mw

I spend some time searching and reading help. But not able to find if this is possible.

I use worksheet only (i.e. not 2D document). I have my display set as

 

I'd like diff(y(x),x) to display as y'(x) in output.

I know I can do this 

PDEtools:-declare(y(x), prime = x);

And that will make diff(y(x),x) display as y'  but I want y'(x). And the same for diff(y(x),x$2) to display as y''(x). And to be clear, y(x) will still display as y(x).  I am mainly interested in making the derivative display a little nicer if possible.

Is there a way to do this?

I am using 2019.1 on windows 10.

 

pde := (diff(u(r, theta), r) + r * diff(u(r, theta), r, r) + diff(u(r, theta), theta, theta) / r ) / r:
iv := u( 1, theta) = 0, u( 3, theta) = theta, u( r, 0) = 10, u( r, Pi/2) = 0:
           Maple 2019 returns a symbolic solution for PDE:
pdsolve([pde, iv], u(r, theta));
   But for the numeric option, it returns a message saying that Maple is unable to handle elliptical PDEs.
pdsolve(pde, {iv}, numeric, time = t, range = 1 .. 3);

Error, (in pdsolve/numeric) unable to handle elliptic PDEs
I found it strange.

Oliveira.

In answers given in 

In https://www.mapleprimes.com/questions/227546-How-To-Make-Odetest-Verify-Dsolve

It shows that odetest() did not verify a solution to ODE becuase solution was using hypergeom special functions. If the solution to the ODE was in integral form, then odetest() will verify it OK.

But what to do if the solution I want to verify is already in hypergoem? If I try odetest() it will fail to verify now. Then I can try to convert the solution to integral form and try again.

But when  using convert(sol,Int) followed by odetest() it did not work.

The solutions I try to verify are hand solutions or book solutions, and not coming from dsolve. 

But some of them are the same solution that comes from dsolve() when not using the useInt option. 

Also, I am doing this all inside a Maple program. It is not an interactive process. So I can't do plots and look at them to decide on anything. So verification must all be implemented in code.

The question is: Why did convert(hand_solution,Int) not give the same result as dsolve(ode,useInt)? Is there another way around this? (May be I am asking for too much in this one based on answers in the above link, So that is OK if not possible. But I really like the solution given when using "useInt" option. Much more clear than otherwise).
 

restart;

ode := diff(y(x), x)*(x^3 + 1)^(2/3) + (1 + y(x)^3)^(2/3) = 0;
sol_int:=dsolve(ode,useInt);
odetest(sol_int,ode); #OK now, since solution in integral form

(diff(y(x), x))*(x^3+1)^(2/3)+(1+y(x)^3)^(2/3) = 0

Int(1/(x^3+1)^(2/3), x)+Intat(1/(_a^3+1)^(2/3), _a = y(x))+_C1 = 0

0

hand_solution:= x*hypergeom([1/3, 2/3], [4/3], -x^3) + y(x)*hypergeom([1/3, 2/3], [4/3], -y(x)^3) + _C1 = 0;
convert(hand_solution,Int); #Why this did not give same result as ABOVE?

x*hypergeom([1/3, 2/3], [4/3], -x^3)+y(x)*hypergeom([1/3, 2/3], [4/3], -y(x)^3)+_C1 = 0

(2/9)*x*Pi*3^(1/2)*(Int(1/(_t1^(1/3)*(1-_t1)^(1/3)*(x^3*_t1+1)^(1/3)), _t1 = 0 .. 1))/GAMMA(2/3)^3+(2/9)*y(x)*Pi*3^(1/2)*(Int(1/(_t1^(1/3)*(1-_t1)^(1/3)*(y(x)^3*_t1+1)^(1/3)), _t1 = 0 .. 1))/GAMMA(2/3)^3+_C1 = 0

odetest(%,ode); #does not give zero

-y(x)^3*(1+y(x)^3)^(2/3)*(Int(_t1^(2/3)/((1-_t1)^(1/3)*(y(x)^3*_t1+1)^(4/3)), _t1 = 0 .. 1))+(x^3+1)^(2/3)*(Int(_t1^(2/3)/((1-_t1)^(1/3)*(x^3*_t1+1)^(4/3)), _t1 = 0 .. 1))*x^3-(x^3+1)^(2/3)*(Int(1/(_t1^(1/3)*(1-_t1)^(1/3)*(x^3*_t1+1)^(1/3)), _t1 = 0 .. 1))+(1+y(x)^3)^(2/3)*(Int(1/(_t1^(1/3)*(1-_t1)^(1/3)*(y(x)^3*_t1+1)^(1/3)), _t1 = 0 .. 1))

 

 

Maple 2019.1

Download 072619_2.mw

 

 

 

It seems that the answer to my question is deleted!!!

So, I again repeat it.

How I can calculate this integral?

I want to calculate integral with the constants Aj]j=1,2.

The Amount of these constants  unknown this stage.

Thanks

INTEGRAL

integral.mw

I am trying to evaluate any which way the integral:

int(exp(-(sqrt(4*x^2+4*y^2+4*z^2)^3)), z = -sqrt(4-x^2-y^2).. sqrt(4-x^2-y^2),y=-sqrt(4-x^2)..sqrt(4-x^2),x=-2..2);

The program just hangs, so i click on 'stop current operation'.

Then I tried:

evalf(Int(exp(-(sqrt(4*x^2+4*y^2+4*z^2)^3)), z = -sqrt(4-x^2-y^2).. sqrt(4-x^2-y^2),y=-sqrt(4-x^2)..sqrt(4-x^2),x=-2..2));

It returns the integral back unevaluated.

It's true that I could use a change of variables, changing to spherical coordinates would be best here. But I would like to know if I did set up the original Cartesian integral correctly. Is there some workaround I can apply to get a numeric answer. I am satisfied with a decimal approximation. Then I can compare to the change of variable result. An exact answer would be even better of course.

THis is another ode which I am not able to get odetest to give zero. Any one knows of a trick to verify this solution? It might be just that the solution is too complicated for odetest to verify?


 

restart;

ode:=diff(y(x),x)*(x^3+1)^(2/3)+(1+y(x)^3)^(2/3) = 0;
sol:=dsolve(ode);

(diff(y(x), x))*(x^3+1)^(2/3)+(1+y(x)^3)^(2/3) = 0

x*hypergeom([1/3, 2/3], [4/3], -x^3)+y(x)*hypergeom([1/3, 2/3], [4/3], -y(x)^3)+_C1 = 0

odetest(sol,ode);

-9*(1+y(x)^3)^(1/3)*(x^3+1)^(2/3)*hypergeom([4/3, 5/3], [7/3], -x^3)*x^3*GAMMA(2/3)*(-y(x)^3)^(1/6)/(9*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*y(x)^3*(-y(x)^3)^(1/6)*(1+y(x)^3)^(1/3)*GAMMA(2/3)-4*Pi*3^(1/2)*LegendreP(-1/3, -1/3, -y(x)^3/(1+y(x)^3)+1/(1+y(x)^3)))+9*y(x)^6*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*GAMMA(2/3)*(-y(x)^3)^(1/6)/(9*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*y(x)^3*(-y(x)^3)^(1/6)*(1+y(x)^3)^(1/3)*GAMMA(2/3)-4*Pi*3^(1/2)*LegendreP(-1/3, -1/3, -y(x)^3/(1+y(x)^3)+1/(1+y(x)^3)))+9*y(x)^3*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*GAMMA(2/3)*(-y(x)^3)^(1/6)/(9*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*y(x)^3*(-y(x)^3)^(1/6)*(1+y(x)^3)^(1/3)*GAMMA(2/3)-4*Pi*3^(1/2)*LegendreP(-1/3, -1/3, -y(x)^3/(1+y(x)^3)+1/(1+y(x)^3)))-4*(1+y(x)^3)^(2/3)*Pi*3^(1/2)*LegendreP(-1/3, -1/3, -(y(x)^3-1)/(1+y(x)^3))/(9*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*y(x)^3*(-y(x)^3)^(1/6)*(1+y(x)^3)^(1/3)*GAMMA(2/3)-4*Pi*3^(1/2)*LegendreP(-1/3, -1/3, -y(x)^3/(1+y(x)^3)+1/(1+y(x)^3)))+4*(1+y(x)^3)^(1/3)*(x^3+1)^(1/3)*Pi*3^(1/2)*LegendreP(-1/3, -1/3, -(x^3-1)/(x^3+1))*(-y(x)^3)^(1/6)/((-x^3)^(1/6)*(9*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*y(x)^3*(-y(x)^3)^(1/6)*(1+y(x)^3)^(1/3)*GAMMA(2/3)-4*Pi*3^(1/2)*LegendreP(-1/3, -1/3, -y(x)^3/(1+y(x)^3)+1/(1+y(x)^3))))

simplify(%);

-9*((4/9)*(1+y(x)^3)^(2/3)*Pi*3^(1/2)*LegendreP(-1/3, -1/3, (-y(x)^3+1)/(1+y(x)^3))*(-x^3)^(1/6)+(-(4/9)*(1+y(x)^3)^(1/3)*(x^3+1)^(1/3)*Pi*3^(1/2)*LegendreP(-1/3, -1/3, (-x^3+1)/(x^3+1))+(-x^3)^(1/6)*((-y(x)^6-y(x)^3)*hypergeom([4/3, 5/3], [7/3], -y(x)^3)+x^3*(1+y(x)^3)^(1/3)*hypergeom([4/3, 5/3], [7/3], -x^3)*(x^3+1)^(2/3))*GAMMA(2/3))*(-y(x)^3)^(1/6))/((-x^3)^(1/6)*(9*hypergeom([4/3, 5/3], [7/3], -y(x)^3)*y(x)^3*(-y(x)^3)^(1/6)*(1+y(x)^3)^(1/3)*GAMMA(2/3)-4*Pi*3^(1/2)*LegendreP(-1/3, -1/3, (-y(x)^3+1)/(1+y(x)^3))))

 


 

Download 072619.mw

Maple 2019.1, Physics 395

Download 072619.mw

 

 

pde := diff(u(x, t), x $ 4) = diff(u(x, t), t $ 2);

iv:= subs(L = 100, {u(0, t) = 0, u(L, t) = 0, u(x, 0) = sin(x), D[2](u)(x, 0) = 2*x, D[1, 1](u)(0, t) = 0, D[1, 1](u)(L, t) = 0});

de := pdsolve(pde, iv, numeric):

sa1 := de:-value(output = listprocedure);

sa1:=[x=proc() ... end proc,t=proc() ... end proc,u(x,t)=proc() .. end proc]

    With the above procedure it works, but in the most compact form below it does not work.

pdsolve(pde, iv, numeric,output = listprocedure):

Error, (in pdsolve/numeric/par_hyp) invalid arguments for theta scheme: [output = listprocedure]
 

What is the solution to this error message when trying to run the Maple add-in for Excel?
"The specified module could not be found.  OpenMaple cannot find Maple engine library, maple.dll."

I am running Maple 19.1 and Excel 2016, both 32-bit, in Windows 7.  Excel appears to have successfully installed the Maple add-in.  For example, the Maple add-in icons appear on the Add-ins tab of Excel.  Excel shows the Maple add-in as active in the list of add-ins.  The location of he WMIMPLEX.xla is correct at "C:\Program Files (x86)\Maple\Excel\WMIMPLEX.xla.

Maple support tells me that Maple should add its "bin" folder to the PATH key for excel.exe in the registry key "Computer\HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\App Paths\excel.exe\Path".  This path information, however, is not in my computer's registry.  This missing path information might be an obvious problem.  But what is the correct registry entry so that Excel knows where to find the Maple engine library maple.dll?

I tried adding the path to maple.dll in the environmental path variables of Windows 7, but that approach did not work.  I have uninstalled, rebooted, and reinstalled Maple 2019 (32-bit) several times.  Still the same error message.

Thanks.

with(PDEtools);
pde := diff(c(x, t), x, x) - h*diff(c(x, t), x) = diff(c(x, t), t):

iv := c(0, t) = 0, c(a, t) = 0, c(x, 0) = c0:

de := pdsolve([pde, iv], c(x, t), build);

                         de := ( )

Does anyone know how to solve this PDE?
Thank you,

Oliveira

      

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